msxsl:node-set with default namespace

[MoonStorm]

Hi guys,

Please solve a puzzle I am trying to figure out for some time.
Let's say I have a fragment stored inside a variable, for instance:

<xsl:variable name="layoutSet tings">
<module>
<size>345</size>
<title>whatever </title>
</module>
</xsl:variable>

My standard namespaces look like this:
<xsl:styleshe et version="1.0"
xmlns:xsl="http ://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="ur n:schemas-microsoft-com:xslt">

The node-set function behaves as expected.
The trouble appears when I try to add a default namespace to the
xsl:stylesheet declaration. A
select="msxsl:n ode-set($moduleSett ings)/module/size" will return
nothing.

[David Carlisle]

The node-set function behaves as expected.
The trouble appears when I try to add a default namespace to the
xsl:stylesheet declaration. A

This behaviour is nothing to do with node-set() you will see the same
behaviour if you add a namespace to a source document. unprefixed
element names in XPath 1 _always_ refer to elements in no-namespace.

By adding a default namespace the elements in your variable are now in
that namespace.

Either you have to prefix the names in the Xpath or add xmlns="" to the
xsl:variable so the elements within it are in no-namespace.

David

[Cody Amor]

Yes, you are right. Now it works.

However, from what I understand and tested, I cannot prefix the names in
xpath with the default namespace. If I choose this option, I have to
declare a new namespace and include the elements in that namespace. Am I
right?

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[David Carlisle]

Cody Amor <mo*******@gmai l.com> writes:

Yes, you are right. Now it works.

However, from what I understand and tested, I cannot prefix the names in
xpath with the default namespace. If I choose this option, I have to
declare a new namespace and include the elements in that namespace. Am I
right?

you need to declare the same namespace twice with and without a prefix

<xsl:styleshe et xmlns="wibble" xmlns="w:wibble " ....

Now a literal result element of
<foo> is foo in wibble namepsace (as is <w:foo> ) and an Xpath of
//x:foo will find foo in the wibble namespace (whether or not it was
prefixed in the source)

David

[Cody Amor]

Brrrrr, yes, it works. Thanks a lot.
Hopefully the standard will evolve in a more friendly way.
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[David Carlisle]

Cody Amor <mo*******@gmai l.com> writes:

Brrrrr, yes, it works. Thanks a lot.
Hopefully the standard will evolve in a more friendly way.

Yes it will, XSLT2 draft (which is already implemented in, eg, saxon8.x)
allows you to specify a default namespace for element names in Xpath
expressions, so you could keep everything unprefixed, and just use the
default namespace, then also declare that unprefixed xpath names refer
to that.

http://www.w3.org/TR/xslt20/#unprefixed-qnames

David